Question 121018

Start with the given system

{{{3x+2y=-5}}}
{{{x=2y-7}}}




{{{3(2y-7)+2y=-5}}}  Plug in {{{x=2y-7}}} into the first equation. In other words, replace each {{{x}}} with {{{2y-7}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{6y-21+2y=-5}}} Distribute



{{{8y-21=-5}}} Combine like terms on the left side



{{{8y=-5+21}}}Add 21 to both sides



{{{8y=16}}} Combine like terms on the right side



{{{y=(16)/(8)}}} Divide both sides by 8 to isolate y




{{{y=2}}} Divide





Now that we know that {{{y=2}}}, we can plug this into {{{x=2y-7}}} to find {{{x}}}




{{{x=2(2)-7}}} Substitute {{{2}}} for each {{{y}}}



{{{x=-3}}} Simplify



So our answer is {{{x=-3}}} and {{{y=2}}} which also looks like *[Tex \LARGE \left(-3,2\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(-3,2\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, (-5-3x)/2, (x+7)/2) }}} Graph of {{{3x+2y=-5}}} (red) and {{{x=2y-7}}} (green)